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If M is the product of all positive integers greater than 59 and less

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Bunuel
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If M is the product of all positive integers greater than 59 and less [#permalink] New post   01 May 2021, 06:15
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If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer?

A. 5
B. 7
C. 9
D. 11
E. 13
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Bunuel
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Re: If M is the product of all positive integers greater than 59 and less [#permalink] New post   09 May 2021, 01:15
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Bunuel wrote:
If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer?

A. 5
B. 7
C. 9
D. 11
E. 13


We need to find the power of of 6 in 60*61*62*63*64*65*66*67*68*69*70.

6 = 2*3.

Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70, than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70:

One 3 in 60;
Two 3's in 63 (63 = 3^2*7);
One 3 in 66;
One 3 in 69.

Total of five 3's.

Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5.

Answer: A.

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Re: If M is the product of all positive integers greater than 59 and less [#permalink] New post   01 May 2021, 08:45
Asked: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer?

M = 70!/59!

Greatest power of 3 in 70! = 23 + 7 + 2 = 32
Greatest power of 3 in 59! = 19 + 6 + 2 = 27
Greatest power of 3 in M = 70!/59! = 5
Greatest power of 6 in M = 5

IMO A
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Re: If M is the product of all positive integers greater than 59 and less [#permalink] New post   01 May 2021, 08:46
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What we have is 70! - 59! and this is being divided by 6^n

We take the greatest prime in 6 which 3 and see how many times it goes into each of the factorials using the formula n/p + n/p^2 + n/p^3....+ n/p^k where n is the factorial we are dealing with and p is the largest prime (3)

70/3=23 , 70/9=7 , 70/27=2
total = 32

59/3=19 , 59/9=6 , 59/27=2
total = 27

32 - 27 =5
Answer A
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Re: If M is the product of all positive integers greater than 59 and less [#permalink] New post   01 May 2021, 08:50
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Asked: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer?

M = 60*61*62*63*64*65*66*67*68*69*70 = (2^2*3*5)*61*(2*31)*(3*3*7)*(2^6)*(5*13)*(2*3*11)*67*(2*2*17)*(3*23)*(2*5*7) = 2^13*3^5*Integer = 6^5*Integer
The greatest integer n for which \(\frac{M}{6^n}\) is an integer = 5

IMO A
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Re: If M is the product of all positive integers greater than 59 and less [#permalink] New post   27 Mar 2024, 00:42
Nidzo wrote:
What we have is 70! - 59! and this is being divided by 6^n

We take the greatest prime in 6 which 3 and see how many times it goes into each of the factorials using the formula n/p + n/p^2 + n/p^3....+ n/p^k where n is the factorial we are dealing with and p is the largest prime (3)

70/3=23 , 70/9=7 , 70/27=2
total = 32

59/3=19 , 59/9=6 , 59/27=2
total = 27

32 - 27 =5
Answer A




Why is it 59/3 instead of 60/3 when the question says number greater than 59?
Thanks!

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Re: If M is the product of all positive integers greater than 59 and less [#permalink]
27 Mar 2024, 00:42
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